Question

In a parallelogram ABCD, E and F are the midpoints of the sides AB and CD respectively .show that the line segments AF and EC trisect the diagonal BD

Answer 1

given ABCD is a parallelogram

Hence AB || CD

⇒ AE || FC

Also AB = CD (Opposite sides of parallelogram ABCD)

⇒ AE = FC (Since E and F are midpoints of AB and CD)

In quadrilateral AECF, one pair of opposite sides are equal and parallel.

∴ AECF is a parallelogram. ⇒ AF || EC (Since opposite sides of a parallelogram are parallel)

In ΔDPC, F is the midpoint of DC and FQ || CP


Hence Q is the midpoint of DQ by converse of midpoint theorem.

⇒ DQ = PQ → (1) Similarly, in ΔAQB, E is the midpoint of AB and EP || AQ

Hence P is the midpoint of DQ by converse of midpoint theorem. ⇒ BP = PQ → (2)

From equations (1) and (2), we get BP = PQ = DQ

Hence, the line segments AF and EC trisect the diagonal BD of parallelogram ABCD.


I hope it is helped you

Answer 2

Answer:

given ABCD is a parallelogram

Hence AB || CD

⇒ AE || FC

Also AB = CD (Opposite sides of parallelogram ABCD)

⇒ AE = FC (Since E and F are midpoints of AB and CD)

In quadrilateral AECF, one pair of opposite sides are equal and parallel.

∴ AECF is a parallelogram. ⇒ AF || EC (Since opposite sides of a parallelogram are parallel)

In ΔDPC, F is the midpoint of DC and FQ || CP

Hence Q is the midpoint of DQ by converse of midpoint theorem.

⇒ DQ = PQ → (1) Similarly, in ΔAQB, E is the midpoint of AB and EP || AQ

Hence P is the midpoint of DQ by converse of midpoint theorem. ⇒ BP = PQ → (2)

From equations (1) and (2), we get BP = PQ = DQ

Hence, the line segments AF and EC trisect the diagonal BD of parallelogram ABCD.

Read more on Brainly.in - https://brainly.in/question/6251645#readmore

Step-by-step explanation: